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In a particular reference frame with coordinates $x^\mu$, we can define a current density 4-vector $J^\mu=(c\rho,\vec{J})$ where $\rho$ is the charge density and $\vec{J}$ is the current density.

The continuity equation in this frame is then $$\frac{\partial J^\mu}{\partial x^\mu}=0.$$

How can it be shown that this equation holds in a Lorentz boosted frame with coordinates $x'^\mu=\Lambda^\mu_\nu x ^\nu$, i.e. $$\frac{\partial J'^\mu}{\partial x'^\mu}=0$$ is true?

TaeNyFan
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2 Answers2

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From $J'^\mu=\Lambda^\mu_\nu J^\nu$ take derivative on both sides $$\frac{\partial J'^\mu}{\partial x'^\mu}=\Lambda^\mu_\nu \frac{\partial J^\nu}{\partial x'^\mu}$$ Note that $\frac{\partial}{\partial x'^\mu}=\Lambda^\sigma_\mu\frac{\partial}{\partial x^\sigma}$ Then we have $$\frac{\partial J'^\mu}{\partial x'^\mu}=\Lambda^\mu_\nu \frac{\partial J^\nu}{\partial x'^\mu}=\Lambda^\mu_\nu\Lambda^\sigma_\mu\frac{\partial J^\nu}{\partial x^\sigma}=\delta^\sigma_\nu\frac{\partial J^\nu}{\partial x^\sigma}=\frac{\partial J^\nu}{\partial x^\nu}=0$$

Not even trivial
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Generally \begin{align} &\texttt{The 4-gradient contravariant vector operator} \nonumber\\ & \qquad \qquad \qquad \partial^{\mu}\boldsymbol{\equiv}\dfrac{\partial }{\partial x_{\mu}}\boldsymbol{=}\left(\dfrac{\partial }{\partial x^{0}}\,,\boldsymbol{-\nabla}\right) \tag{01a}\label{01a}\\ &\texttt{The 4-gradient covariant vector operator} \nonumber\\ &\qquad \qquad \qquad \partial_{\mu}\boldsymbol{\equiv}\dfrac{\partial }{\partial x^{\mu}}\boldsymbol{=}\left(\dfrac{\partial }{\partial x^{0}}\,,\boldsymbol{+\nabla}\right) \tag{01b}\label{01b} \end{align} The 4-divergence of a 4-vector $A^{\mu}\boldsymbol{=}\left(A^{0},\mathbf{A}\right),A_{\mu}\boldsymbol{=}\left(A^{0},\boldsymbol{-}\mathbf{A}\right) $ is the invariant \begin{equation} \partial^{\mu}A_{\mu}\boldsymbol{=}\partial_{\mu}A^{\mu}\boldsymbol{=}\dfrac{\partial A^{0} }{\partial x^{0}}\boldsymbol{+\nabla\cdot}\mathbf{A} \tag{02}\label{02} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

Note : May be in the future you ask yourself if the converse of \eqref{02} is true. No, it's not : if $A^{\mu}\boldsymbol{=}\left(A^{0},\mathbf{A}\right)$ is a 4-dimensional quantity then \begin{equation} \partial_{\mu}A^{\mu}\boldsymbol{=}\texttt{invariant}\quad \boldsymbol{=\!\ne\!\Rightarrow}\quad A^{\mu}\boldsymbol{=}\texttt{contravariant Lorentz 4-vector} \tag{03}\label{03} \end{equation}

$-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!$

Counter example

Consider that $Y^{\mu}\left(\mathbf{x},t\right)\boldsymbol{=}\left[\!\!\left[ Y^{0}\left(\mathbf{x},t\right),\mathbf{Y}\left(\mathbf{x},t\right)\right]\!\!\right]$ is a (contravariant) 4-vector function of the space-time coordinates $\left(\mathbf{x},t\right)$. According to equation \eqref{02} its 4-divergence is invariant \begin{equation} \partial_{\mu}Y^{\mu}\boldsymbol{=}\dfrac{\partial Y^{0} }{\partial x^{0}}\boldsymbol{+\nabla\cdot}\mathbf{Y}\boldsymbol{=}\texttt{invariant} \tag{04}\label{04} \end{equation} Now consider a 4-dimensional vector $\rm a^{\mu}\boldsymbol{=}\left(a^{0},\mathbf{a}\right)\boldsymbol{=}\left(a^{0},a^{1},a^{2},a^{3}\right)$ with components 4 arbitrary real numbers constants, that is not depending on the space-time coordinates $\left(\mathbf{x},t\right)$ and form the 4-dimensional vector \begin{equation} A^{\mu}\boldsymbol{=}Y^{\mu}\boldsymbol{+}\rm a^{\mu} \tag{05}\label{05} \end{equation} Then \begin{equation} \partial_{\mu}A^{\mu}\boldsymbol{=}\partial_{\mu}Y^{\mu}\boldsymbol{+}\overbrace{\partial_{\mu}\rm a^{\mu}}^{0}\boldsymbol{=}\partial_{\mu}Y^{\mu}\boldsymbol{=}\texttt{invariant} \tag{06}\label{06} \end{equation} But $A^{\mu}$ as defined by equation \eqref{05} in not a 4-vector due mainly to the arbitrariness of $\rm a^{\mu}$.

Frobenius
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    To show that the 4 divergence of a 4 vector is invariant under coordinate transformations, we still have to use Frank's answer, right? – TaeNyFan Dec 12 '20 at 05:14
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    @TaeNyFan : Yes, precisely. The proof is given in Frank's answer that's why I didn't post the same one. – Frobenius Dec 12 '20 at 08:10
  • @Frobenius Could you give a counter example to your note? I remember this proof I saw a while ago (I might be wrong, I haven't checked) but it said something like: if $B_\mu A^\mu$ is a Lorentz invariant, and $B_\mu$ transforms like a four-vector, then $A^\mu$ must be a four-vector. Are you saying that that doesn't hold here since in our case the "$B_\mu$" is an operator $\partial_\mu$? Or is the "theorem" I've stated wrong? – Philip Dec 13 '20 at 08:09
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    @Philip : I think you refer to this post here Proving an object is a 4-vector given its inner product with a 4-vector is a scalar. Note that the condition $B_\mu A^\mu$ to be a Lorentz invariant with one only specific 4-vector $B_\mu$ it's not sufficient for $A^\mu$ to be a 4-vector. It's necessary the condition $B_\mu A^\mu$ to be a Lorentz invariant with any arbitrary 4-vector $B_\mu$. I added a Counter example in my answer. – Frobenius Dec 13 '20 at 10:14
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    Ah! Very interesting, thank you! – Philip Dec 13 '20 at 10:25